P->Q 2 years ago simplify into a single fraction and write your answer in terms of sin x and cos x: tan x + 1/tan x

1. ZeHanz

Hint: tanx=sinx/cosx, so 1/tanx=cosx/sinx. You now only need to write the sum of these as one fraction!

2. P->Q

so sin^2x + Cos^2x/cosxsinx ?

3. P->Q

anyone?

4. asnaseer

yes - that is correct but you can simplify this further.$\sin^2(x)+\cos^2(x)=?$

5. P->Q

o um (sinx)(sinx) + (cosx)(cosx)/cosxsinx

6. asnaseer

no, the equation I wrote up there is a well know trig identity - look it up in your notes, you must have been taught this at some stage.

7. P->Q

1

8. P->Q

Pythagorean trig identity

9. asnaseer

correct

10. asnaseer

so now make use of that fact to simplify your expression

11. P->Q

hmmm

12. P->Q

ya idk, I just kinda put it out sin^2x + Cos^2x/cosxsinx = 1 im not seeing how this relates it just made it worse,.

13. asnaseer

the identity you need to use is:$\sin^2(x)+\cos^2(x)=1$

14. P->Q

what do u mean, this has nothing to do with this problem lol I mean all I know is that is the Pythagorean tri identity and that you can move cos or sin and lets say u move sin it goes to the right and u have cos^2(x) = plus or minus radical 1-sin^2(x)

15. asnaseer

Look at the steps you took in this problem:\begin{align} \tan(x)+\frac{1}{\tan(x)}&=\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}\\ &=\frac{\sin^2(x)+\cos^2(x)}{\cos(x)\sin(x)} \end{align}now notice that the numerator contains the terms: $$\sin^2(x)+\cos^2(x)$$, and you know that this identity always equals 1. Therefore you can replace $$\sin^2(x)+\cos^2(x)$$ by 1.